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Proof step 6 from proof sub-01 (previous step)

¬(0 ≤ i0 ∧ 0 ≤ j), j ≤ 0, i0 + j ≤ 0, i0 ≤ 0 ⊦ i→n(i0) = i→n(i0 + j) - i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i0 + j ≤ 0, j ≤ 0, i0 < 0 ⊦ i→n(i0) = i→n(i0 + j) - i→n(j), i0 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero: ⊦ m - m = 0; from specification nat%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs